from math 2220 class 13pi.math.cornell.edu/~back/m222_f14/slides/sep26_v2c.pdfsep. 26, 2014 from...

From Math2220 Class 13

V2c

Prelim andHW

AbsoluteExtrema

1 VariableTaylor Series

MultivariableTaylorPolynomials

MultivariableTaylorPictures

Higher PartialDerivatives

A Counterex-ample

Higher Partialsin PolarCordinates

GradientEstimate fromGraph

Method ofCharacteristics

From Math 2220 Class 13

Dr. Allen Back

Sep. 26, 2014


V2c

Prelim andHW

AbsoluteExtrema





A Counterex-ample




Prelim and HW

Your next homework on 3.2 and 3.3 will not be due soonerthan Mon 10 days from now.


V2c

Prelim andHW

AbsoluteExtrema





A Counterex-ample




Prelim and HW

No homework quiz today.


V2c

Prelim andHW

AbsoluteExtrema





A Counterex-ample




Prelim and HW

Prelim 1 next Tues at 7:30 pm in Malott 251 will include 3.1and the local part of 3.3 but not 3.2. (Chapter 2 as well.)We can spend as much of Monday on your review questions asyou like.


V2c

Prelim andHW

AbsoluteExtrema





A Counterex-ample




Prelim and HW

The exam has 6 questions and is not very intricatecomputationally. Question 6 is a True/False question (not toohard) with the following directions:


V2c

Prelim andHW

AbsoluteExtrema





A Counterex-ample




Prelim and HW

Problem 6) (15 Points) For the questions below, please justwrite the complete word TRUE or FALSE as your answer. Forthis question ONLY, no explanations are expected.(If a statement is sometimes true but not always, your answerwould be FALSE.)

You will receive 3 points for a correct answer, 1 point for ablank and −1 point for a wrong answer.

(Thus all 5 parts blank would give you 5 out of 15 while all fivewrong will lower your total on the rest of the exam by 5.)

(The exam is 11 pages with space below the questions to writeyour answers.)


V2c

Prelim andHW

AbsoluteExtrema





A Counterex-ample




Prelim and HW

In addition to my usual Monday office hours (Evan has sometoo), I will have office hours

Fri 9/26 2:15-3:15

Tues 9/30 1-2:30

Also I’m willing to talk with anyone who likes rightafter class today though I do have an appt withEvan for 2-2:15.


V2c

Prelim andHW

AbsoluteExtrema





A Counterex-ample




Absolute Extrema

Closed Set contains all its boundary points.


V2c

Prelim andHW

AbsoluteExtrema





A Counterex-ample




Absolute Extrema

Bounded Set lies inside some single ball of some radius.


V2c

Prelim andHW

AbsoluteExtrema





A Counterex-ample




Absolute Extrema

Compact Set both closed and bounded.


V2c

Prelim andHW

AbsoluteExtrema





A Counterex-ample




Absolute Extrema

Theorem A continuous functions whose domain is a compactset always attains (i.e. “has”) an absolute maximum and anabsolute minimum.


V2c

Prelim andHW

AbsoluteExtrema





A Counterex-ample




Absolute Extrema

Counterexamples when not compact:


V2c

Prelim andHW

AbsoluteExtrema





A Counterex-ample




Absolute Extrema

Procedure for finding an absolute extremum for a cont. fcn. ona compact domain:

1 Above theorem guarantees existence since the domain iscompact.

2 Look for critical points (in the interior of the domain.)

3 Investigate the boundary either using lower dimensionalcalculus or Lagrange multipliers.

4 Compare values at all candidates to find the absolute maxand min.


V2c

Prelim andHW

AbsoluteExtrema





A Counterex-ample




Absolute Extrema

Find the absolute max and absolute min of

f (x , y) = xy(2− x − y)

on the square |x | ≤ 1, |y | ≤ 1.


V2c

Prelim andHW

AbsoluteExtrema





A Counterex-ample




Absolute Extrema

Find the absolute max and absolute min of

f (x , y) = xy

on (and inside) the triangle with vertices (2, 0), (10, 0), and(4, 1).


V2c

Prelim andHW

AbsoluteExtrema





A Counterex-ample




1 Variable Taylor Series

Multivariable Taylor approximation follows (using the chainrule) easily from the 1-variable case, so we’ll start by reviewinghow the results there work out.


V2c

Prelim andHW

AbsoluteExtrema





A Counterex-ample





Taylor Series of f (x) about x = a:

Σ∞n=0

f n(a)

n!(x − a)n.


V2c

Prelim andHW

AbsoluteExtrema





A Counterex-ample





Partial sums (the k + 1’st) give the k ’th Taylor Polynomial.

Pk(x) = Σkn=0

f n(a)

n!(x − a)n.

(Pk is of degree k but has k + 1 terms because we start with0.)


V2c

Prelim andHW

AbsoluteExtrema





A Counterex-ample





So P1(x) is the linear approximation to f (x) at x = a.


V2c

Prelim andHW

AbsoluteExtrema





A Counterex-ample





Examples:

(a)ex about x = 0 (b)ex about x = 2


V2c

Prelim andHW

AbsoluteExtrema





A Counterex-ample





Where does the formula for Pk come from heuristically?


V2c

Prelim andHW

AbsoluteExtrema





A Counterex-ample





Where does the formula for Pk come from heuristically?The k’th Taylor polynomial Pk(x) is the unique polynomial ofdegree k whose value, derivative, 2nd derivative, . . . k’thderivative all agree with those of f (x) at x = a.


V2c

Prelim andHW

AbsoluteExtrema





A Counterex-ample





Where does the formula for Pk come from heuristically?Or integration by parts: u = f ′(x), v ′ = 1, v = (x − b)

f (b) = f (a) +

∫ b

af ′(x) dx

f (b) = f (a) +(

(x − b)f ′(x)∣∣ba−

∫ b

a(x − b)f ′′(x) dx

)f (b) = f (a) + (b − a)f ′(a)+

∫ b

a(b − x)f ′′(x) dx

The last term is one form of the “remainder” in approximatingf (b) by the first Taylor polynomial (aka linear approximant)P1(b).


V2c

Prelim andHW

AbsoluteExtrema





A Counterex-ample





Where does the formula for Pk come from heuristically?

Or integration by parts: u = f ′′(x), v ′ = (b − x), v = −(b−x)2

2

f (b) = f (a) + (b − a)f ′(a)+

∫ b

a(b − x)f ′′(x) dx

)f (b) = f (a) + (b − a)f ′(a)+

(−(b − x)2

2f ′′(x)

∣∣∣∣ba

−∫ b

a

−(b − x)2

2f ′′′(x) dx

)


V2c

Prelim andHW

AbsoluteExtrema





A Counterex-ample





Where does the formula for Pk come from heuristically?Or integration by parts:

f (b) = f (a) + (b − a)f ′(a)+

(−(b − x)2

2f ′′(x)

∣∣∣∣ba

−∫ b

a

−(b − x)2

2f ′′′(x) dx

)f (b) = f (a) + (b − a)f ′(a)+

(b − a)2

2f ′′(a)

+

∫ b

a

(b − x)2

2f ′′′(x) dx


V2c

Prelim andHW

AbsoluteExtrema





A Counterex-ample





The above “integral formula” for the remainder can be turnedinto something simpler to remember but sufficient for mostapplications:The next Taylor term except with the higher derivativeevaluated at some unknown point c between a and b.


V2c

Prelim andHW

AbsoluteExtrema





A Counterex-ample





Taylor’s Theorem with RemainderIf f (x), f ”(x), . . . f (k)(x) all exist and are continuous on [a, b]and f (k+1)(x) exists on (a, b), then there is a c ∈ (a, b) so that

f (b) = Pk(b) + Rk(b)

where Pk is the k’th Taylor polynomial of f about x = a and

Rk(b) =f k+1(c)

(k + 1)!(b − a)k+1.

Please remember this!


V2c

Prelim andHW

AbsoluteExtrema





A Counterex-ample





Example: Accuracy of sin x byP1(x) (or P3(x)) (about x = 0)for |x | < .1? Or x < .01?


V2c

Prelim andHW

AbsoluteExtrema





A Counterex-ample





f (x) = 11+x+x2 about x = 1.

Approximation properties of different Taylor polynomials nearx = 1.


V2c

Prelim andHW

AbsoluteExtrema





A Counterex-ample





Taylor polynomials of f (x) = 11+x+x2 about x = 1.

Typically higher order approximate well for a greater distance.

(Bad behavior beyond the radius of convergence.)


V2c

Prelim andHW

AbsoluteExtrema





A Counterex-ample





Show that the Maclaurin series of sin x converges to sin x forall x .


V2c

Prelim andHW

AbsoluteExtrema





A Counterex-ample





Show that the Maclaurin series of sin x converges to sin x forall x .This is different than saying the radius of convergence is ∞.Some functions, e.g.

f (x) =

{0 if x ≤ 0

e−1x if x > 0

have a Taylor series which always converges, but not to f (x)even near 0. In this case, every derivative is 0 at x = 0 and theMaclaurin series is identically 0!


V2c

Prelim andHW

AbsoluteExtrema





A Counterex-ample





Show that the Maclaurin series of sin x converges to sin x forall x .The point here is that all derivatives are bounded functionstaking values between −1 and +1. So you can show theremainder Rn(x) goes to 0 as n→∞.Taylor polynomials of f (x) = 1

1+x+x2 about x = 1.Typically higher order approximate well for a greater distance.

(Bad behavior beyond the radius of convergence.)


V2c

Prelim andHW

AbsoluteExtrema





A Counterex-ample





It is more advanced than our course, but there is an amazingconnection between when Taylor series converge to a functionand differentiability of f(x) for x viewed as a complex variable.


V2c

Prelim andHW

AbsoluteExtrema





A Counterex-ample





It is more advanced than our course, but there is an amazingconnection between when Taylor series converge to a functionand differentiability of f(x) for x viewed as a complex variable.For example, consider the MacLaurin series of f (x) = 1

1+x2 .

This fcn. is complex differentiable when x 6= ±i = ±√−1. And

if one identifies the complex number a + bi with the point(a, b) ∈ R2, the nearest bad point of ±i ∼ (0,±1) is distance 1from 0 = 0 + 0i ∼ (0, 0). So from this advanced point of view,the radius of convergence is 1!


V2c

Prelim andHW

AbsoluteExtrema





A Counterex-ample





Similarly, consider the above picture for f (x) = 11+x+x2 about

x = 1. This fcn. is bad when 1 + x + x2 = 0, or x = −1±i√

32 .

The distance from 1 + 0i ∼ (1, 0) to −(12 ,√

32 ) is

√3, so from

this advanced point of view, that is the radius of convergencefor this Taylor series.


V2c

Prelim andHW

AbsoluteExtrema





A Counterex-ample




Multivariable Taylor Polynomials

Given f : U ⊂ Rn → R, to find the formula for the k’th Taylorpolynomial Pk of a C k function about the pointp0 = (a1, a2, . . . , an) ∈ U, we consider the 1-variable functiong(t) defined by

g(t) = f (p0 + th)

where h = (h1, h2, . . . , hn) ∈ Rn.If we are interested in relating the Taylor polynomial Pk to thebehavior of f at the point p = (x1, x2, . . . , xn) ∈ U we mightchoose

h = (x1 − a1, x2 − a2, . . . , xn − an)

as long as the entire line segment between p0 and p lies in U.


V2c

Prelim andHW

AbsoluteExtrema





A Counterex-ample





g(t) = f (p0 + th)


V2c

Prelim andHW

AbsoluteExtrema





A Counterex-ample





Problem: Find the first and second Taylor polynomials of

f (x , y) = cos (x + 3y) + sin x

about (x , y) = (π

2, π).


V2c

Prelim andHW

AbsoluteExtrema





A Counterex-ample






f (x , y) = cos (x + 3y) + sin x

about (x , y) = (π

2, π).

Estimate the error in approximating f by P1 for |x − π

2| < .1

and |y − π| < .1.


V2c

Prelim andHW

AbsoluteExtrema





A Counterex-ample






f (x , y) = cos (x + 3y) + sin x

about (x , y) = (π

2, π).

Estimate the error in approximating f by P1 for |x − π

2| < .01

and |y − π| < .01.


V2c

Prelim andHW

AbsoluteExtrema





A Counterex-ample






f (x , y) = cos (x + 3y) + sin x

about (x , y) = (π

2, π).

Find an ε so that the error in approximating f by P1 for

|x − π

2| < ε and |y − π| < ε is at most .01.

Or at most 10−6.


V2c

Prelim andHW

AbsoluteExtrema





A Counterex-ample




Multivariable Taylor Pictures

Graph of f (x , y) = 1x+y2 for .5 ≤ x , y ≤ 1.5


V2c

Prelim andHW

AbsoluteExtrema





A Counterex-ample





P2 about (1, 1) for f (x , y) = 1x+y2 for .5 ≤ x , y ≤ 1.5


V2c

Prelim andHW

AbsoluteExtrema





A Counterex-ample





Both superimposed for .5 ≤ x , y ≤ 1.5


V2c

Prelim andHW

AbsoluteExtrema





A Counterex-ample





Error in using P2 for f for .5 ≤ x , y ≤ 1.5


V2c

Prelim andHW

AbsoluteExtrema





A Counterex-ample





Error in using P2 for f for .75 ≤ x , y ≤ 1.25


V2c

Prelim andHW

AbsoluteExtrema





A Counterex-ample





Error in using P2 for f for 0 ≤ x , y ≤ 3


V2c

Prelim andHW

AbsoluteExtrema





A Counterex-ample




Higher Partial Derivatives

A typical application of mixed partials arises out of yourhomework problem 3.1 #22 where you show that the waveequation

∂2w

∂x2=∂2w

∂y2

becomes after a change of coordinates

x = u + v , y + u − v

∂2w

∂u∂v= 0


V2c

Prelim andHW

AbsoluteExtrema





A Counterex-ample




Higher Partial Derivatives

A typical application of mixed partials arises out of yourhomework problem 3.1 #22 where you show that the waveequation

∂2w

∂x2=∂2w

∂y2

x = u + v , y + u − v∂2w

∂u∂v= 0

implying the general solution

w = f (u) + g(v) = f (x − y

2) + g(

x + y

2)

for any C 2 functions f and g .


V2c

Prelim andHW

AbsoluteExtrema





A Counterex-ample




A Counterexample

When f is not C 2, fxy need not equal fyx even when both exist.


V2c

Prelim andHW

AbsoluteExtrema





A Counterex-ample




A Counterexample

An example illustrating this is

f (x , y) =

{xy(x2−y2)

x2+y2 if (x , y) 6= (0, 0)

0 if (x , y) = (0, 0)

This function


V2c

Prelim andHW

AbsoluteExtrema





A Counterex-ample




A Counterexample


f (x , y) =

{xy(x2−y2)

x2+y2 if (x , y) 6= (0, 0)

0 if (x , y) = (0, 0)

This functionis clearly C 2 for (x , y) 6= (0, 0) since it is a rational function(quotient of polynomials) with nonzero denominator.


V2c

Prelim andHW

AbsoluteExtrema





A Counterex-ample




A Counterexample


f (x , y) =

{xy(x2−y2)

x2+y2 if (x , y) 6= (0, 0)

0 if (x , y) = (0, 0)

This functionsatisfies f (x , y) = −f (y , x) implying

fxy = fyx

whenever x = y . (A good chain rule exercise . . . )


V2c

Prelim andHW

AbsoluteExtrema





A Counterex-ample




A Counterexample


f (x , y) =

{xy(x2−y2)

x2+y2 if (x , y) 6= (0, 0)

0 if (x , y) = (0, 0)

This functionsatisfies fx(0, 0) = 0 and fy (0, 0) = 0 since f vanishes along theaxes.


V2c

Prelim andHW

AbsoluteExtrema





A Counterex-ample




A Counterexample


f (x , y) =

{xy(x2−y2)

x2+y2 if (x , y) 6= (0, 0)

0 if (x , y) = (0, 0)

This function

satisfies fx =y(x2 − y2)

x2 + y2+

2x2y

x2 + y2− 2x2y(x2 − y2)

(x2 + y2)2.

implying

fx(0, y) = −y (the last two terms vanish)

and∂2f

∂y∂x(0, 0) = −1.


V2c

Prelim andHW

AbsoluteExtrema





A Counterex-ample




A Counterexample


f (x , y) =

{xy(x2−y2)

x2+y2 if (x , y) 6= (0, 0)

0 if (x , y) = (0, 0)

This function

fx(0, y) = −y (the last two terms vanish)

and∂2f

∂y∂x(0, 0) = −1.

By fxy (x , x) = −fyx(x , x) (mentioned above) or directly we see

∂2f

∂x∂y(0, 0) = 1

and the mixed partials really differ!


V2c

Prelim andHW

AbsoluteExtrema





A Counterex-ample




Higher Partials in Polar Cordinates

Let x = r cos θ, y = r sin θ. Given f (x , y), express

∂2f

∂r2

in terms of partials of f with respect to x and y .


V2c

Prelim andHW

AbsoluteExtrema





A Counterex-ample





Really there are two functions here; namely f (x , y) and

g(r , θ) = f (r cos θ, r sin θ).

But many applied fields routinely use the same letter f for bothfunctions.


V2c

Prelim andHW

AbsoluteExtrema





A Counterex-ample





The key step is realizing that the task (via the chain rule) of

relating∂f

∂rto

∂f

∂xand

∂f

∂yis just like the task of doing the

analagous thing with expressions like

∂

∂r

(∂f

∂x

).

(The right hand side “operator notation” means partially

differentiate the function∂f

∂xof (x , y) with respect to r .)


V2c

Prelim andHW

AbsoluteExtrema





A Counterex-ample





x = r cos θ y = r sin θ

∂f

∂r=

∂f

∂x

∂x

∂r+∂f

∂y

∂y

∂r

=∂f

∂xcos θ +

∂f

∂ysin θ

∂

∂r

(∂f

∂x

)=

(∂

∂x

(∂f

∂x

))∂x

∂r+

(∂

∂y

(∂f

∂x

))∂y

∂r

=

(∂2f

∂x2

)cos θ +

(∂2f

∂y∂x

)sin θ


V2c

Prelim andHW

AbsoluteExtrema





A Counterex-ample





A similar computation with∂2f

∂θ2would show that the laplacian

∂2f

∂x2+∂2f

∂y2

is given in polar coordinates by

∂2f

∂r2+

1

r

∂f

∂r+

1

r2

(∂2f

∂θ2

).


V2c

Prelim andHW

AbsoluteExtrema





A Counterex-ample




Can you estimate the value of the gradient?


V2c

Prelim andHW

AbsoluteExtrema





A Counterex-ample




Method of Characteristics

Problem 33 on page 146 from your homework asks you to show

using the chain rule that if z = f (y

x) for a 1-variable

differentiable function f , then

x∂z

∂x+ y

∂z

∂y= 0.


V2c

Prelim andHW

AbsoluteExtrema





A Counterex-ample





Problem 33 on page 146 from your homework asks you to show

using the chain rule that if z = f (y

x) for a 1-variable

differentiable function f , then

x∂z

∂x+ y

∂z

∂y= 0.

You don’t need it to do this homework problem, but a naturalquestion is

Does x∂z

∂x+ y

∂z

∂y= 0 imply z = f (

y

x)

for some 1-variable differentiable function f ?


V2c

Prelim andHW

AbsoluteExtrema





A Counterex-ample





The answer is yes, and we can easily show it using what welearned in section 2.6!(i.e. essentially chain rule ideas applied to curves.)


V2c

Prelim andHW

AbsoluteExtrema





A Counterex-ample





The beautiful idea for the equation

a(x , y)∂z

∂x+ b(x , y)

∂z

∂y= c(x , y , z)

is to consider curves (x(t), y(t) called characteristics) satisfying

dx

dt= a(x(t), y(t))

dy

dt= b(x(t), y(t)).


V2c

Prelim andHW

AbsoluteExtrema





A Counterex-ample





Here for

x∂z

∂x+ y

∂z

∂y= 0

we havedx

dt= x

dy

dt= y

with solutions x(t) = x0et and y(t) = y0et .


V2c

Prelim andHW

AbsoluteExtrema





A Counterex-ample





Here for

x∂z

∂x+ y

∂z

∂y= 0

we havedx

dt= x

dy

dt= y

with solutions x(t) = x0et and y(t) = y0et .The key observation is that for a solution z(x , y)

d

dt[z(x(t), y(t))] =

∂z

∂x

dx

dt+

∂z

∂y

dy

dt

=∂z

∂xx+

∂z

∂yy

= 0 !


V2c

Prelim andHW

AbsoluteExtrema





A Counterex-ample





Here for

x∂z

∂x+ y

∂z

∂y= 0

with solutions x(t) = x0et and y(t) = y0et .The key observation is that for a solution z(x , y)

d

dt[z(x(t), y(t))] =

∂z

∂x

dx

dt+

∂z

∂y

dy

dt

=∂z

∂xx+

∂z

∂yy

= 0 !

So z(x , y) is constant along linesy

x= k for any k and thus

z(x , y) = z(1,y

x) = f (

y

x)

as we wanted to show.

from math 2220 class 13pi.math.cornell.edu/~back/m222_f14/slides/sep26_v2c.pdfsep. 26, 2014 from...

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